Integrand size = 27, antiderivative size = 121 \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right ) \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(e+f x)\right ) \tan (e+f x) (d \tan (e+f x))^m}{f (1+m)}+\frac {2 b \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c f (3+2 m)} \]
a*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-tan(f*x+e)^2)*tan(f*x+e)*(d*tan(f* x+e))^m/f/(1+m)+2*b*hypergeom([1, 3/4+1/2*m],[7/4+1/2*m],-tan(f*x+e)^2)*(c *tan(f*x+e))^(3/2)*(d*tan(f*x+e))^m/c/f/(3+2*m)
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.51 \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right ) \, dx=\frac {\left (\left (a-b \sqrt [4]{-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,-\frac {\sqrt {c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )+\left (a+i b \sqrt [4]{-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,-\frac {i \sqrt {c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )+a \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,\frac {i \sqrt {c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )-i b \sqrt [4]{-c^2} \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,\frac {i \sqrt {c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )+a \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,\frac {\sqrt {c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )+b \sqrt [4]{-c^2} \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,\frac {\sqrt {c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )\right ) \tan (e+f x) (d \tan (e+f x))^m}{4 f (1+m)} \]
(((a - b*(-c^2)^(1/4))*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, -(Sqrt[c*T an[e + f*x]]/(-c^2)^(1/4))] + (a + I*b*(-c^2)^(1/4))*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, ((-I)*Sqrt[c*Tan[e + f*x]])/(-c^2)^(1/4)] + a*Hypergeo metric2F1[1, 2*(1 + m), 3 + 2*m, (I*Sqrt[c*Tan[e + f*x]])/(-c^2)^(1/4)] - I*b*(-c^2)^(1/4)*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, (I*Sqrt[c*Tan[e + f*x]])/(-c^2)^(1/4)] + a*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, Sqrt[c *Tan[e + f*x]]/(-c^2)^(1/4)] + b*(-c^2)^(1/4)*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, Sqrt[c*Tan[e + f*x]]/(-c^2)^(1/4)])*Tan[e + f*x]*(d*Tan[e + f *x])^m)/(4*f*(1 + m))
Time = 0.61 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4153, 7267, 30, 2370, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \frac {c \int \frac {(d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )}{\tan ^2(e+f x) c^2+c^2}d(c \tan (e+f x))}{f}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \frac {2 c \int \frac {\sqrt {c \tan (e+f x)} \left (c d \tan ^2(e+f x)\right )^m (a+b c \tan (e+f x))}{c^4 \tan ^4(e+f x)+c^2}d\sqrt {c \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \int \frac {(c \tan (e+f x))^{\frac {1}{2} (2 m+1)} \left (a+b \sqrt {c \tan (e+f x)}\right )}{c^4 \tan ^4(e+f x)+c^2}d\sqrt {c \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 2370 |
\(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \int \left (\frac {a (c \tan (e+f x))^{\frac {1}{2} (2 m+1)}}{c^4 \tan ^4(e+f x)+c^2}+\frac {b (c \tan (e+f x))^{\frac {1}{2} (2 m+2)}}{c^4 \tan ^4(e+f x)+c^2}\right )d\sqrt {c \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \left (\frac {a (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-c^2 \tan ^4(e+f x)\right )}{2 c^2 (m+1)}+\frac {b (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2 m+3),\frac {1}{4} (2 m+7),-c^2 \tan ^4(e+f x)\right )}{c^2 (2 m+3)}\right )}{f}\) |
(2*c*(c*d*Tan[e + f*x]^2)^m*((a*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -(c^2*Tan[e + f*x]^4)]*(c*Tan[e + f*x])^(1 + m))/(2*c^2*(1 + m)) + (b*Hyp ergeometric2F1[1, (3 + 2*m)/4, (7 + 2*m)/4, -(c^2*Tan[e + f*x]^4)]*(c*Tan[ e + f*x])^((3 + 2*m)/2))/(c^2*(3 + 2*m))))/(f*(c*Tan[e + f*x])^m)
3.5.7.3.1 Defintions of rubi rules used
Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & & !IntegerQ[p]
Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[ {v = Sum[(c*x)^(m + ii)*((Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2) )/(c^ii*(a + b*x^n))), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{ a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
\[\int \left (a +b \sqrt {c \tan \left (f x +e \right )}\right ) \left (d \tan \left (f x +e \right )\right )^{m}d x\]
\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right ) \, dx=\int { {\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right ) \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{m} \left (a + b \sqrt {c \tan {\left (e + f x \right )}}\right )\, dx \]
\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right ) \, dx=\int { {\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right ) \, dx=\int { {\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right ) \, dx=\int \left (a+b\,\sqrt {c\,\mathrm {tan}\left (e+f\,x\right )}\right )\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m \,d x \]